I believe (1) is the family described in equation (1), which apparently (as per @Noam's and @Mark's guesses) has a fixed Galois group of order 1344, except for the exceptional values when the polynomial is reducible. I am not at all sure what the significance of the squareness of the discrimant is (I mean, the Galois group is then a subgroup of $A_8,$ but so?)
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Igor RivinApr 26 '12 at 2:15

2

For your second question, Malle in "Multi-parameter Polynomials with Given Galois Group" (citeseerx.ist.psu.edu/viewdoc/…) constructs a 3-parameter family of polynomials with Galois group $PSL(2,7)=PSL(3,2)$ (section 4). Then in section 6, he explains that, essentially by taking generic square roots you can construct a 4-parameter family with Galois group $2^3.PSL(3,2)$, which is your group. Then by fixing some of the parameters, you might be able to kill of the dependence on them in all terms but $x^2$.
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Tim DokchitserApr 26 '12 at 8:49

3 Answers
3

Maple gives Galois group $H$ of your polynomial $f=f_n$ as the subgroup of $S_8$ generated by these permutations: $$(1 2)(5 6), (1 2 3)(4 6 5), (1 2 6 3 4 5 7), (1 8)(2 3)(4 5)(6 7), (2 8)(1 3)(4 6)(5 7), (4 8)(1 5)(2 6)(3 7).$$ That can be easily proved by using the standard technique of Galois theory assuming that $n$ is such that your polynomial is irreducible (I think that the only exceptional values are 17 and 145 as in Alastair's answer). See Van der Waerden's book, for example: consider the polynomial in $9$ variables $$g(x,x_1,...,x_8)=\prod_{\sigma\in S_8} (x-x_{\sigma 1}a_1-...-x_{\sigma 8}a_8) $$
where $a_1,...,a_8$ are the roots of your polynomial $f=f_n$, $\sigma$ runs over all permutations of $S_8$. The coefficients of $g$ are symmetric polynomials in $a_i$, hence polynomials in the coefficients of $f_n$. Now show that the products of factors corresponding to permutations from $H$ form an irreducible factor of $g$. Hence $H$ is the Galois group of $f_n$ (again, assuming $f_n$ is irreducible).

Thanks, Mark. I believe this takes care of the first part of the question. But there was a second part, namely: "how easy is it to find, for the same generic Galois group as above, another family where all coefficients remain constant except one?". Kluener's database gives more than a thousand examples for G1344, but this was the only family, among many I found, with a form like that.
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Tito Piezas IIIApr 26 '12 at 3:10

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@Tito: I have no idea about the second part. In fact I would like to know how you got the first family (and also why is it interesting, what is the role of 1344, are there other groups except this, $S_n$ and $A_n$ with this property, etc.?)
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Mark SapirApr 26 '12 at 7:31

@Mark: My interest is really more on solvable groups and, for the octic, especially those with order divisible by 7, such as Schein's nice $x^8-x^7+29x^2+29=0$ which can be solved using the 29th root of unity. While browsing Kluener's database for the non-solvable group 8T48, it was easy to notice a series of polynomials that differed only by a single coefficient. Further experimentation with other values of $n$ confirmed that certain expressions of the octic roots were 7th deg algebraic numbers. As to your question, it seems the family $x^8-x^6-2x^5+2x^3+x^2+n^2=0$ has group $S_8$.
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Tito Piezas IIIApr 26 '12 at 15:16

Exceptions for what? This is rather cryptic.
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Igor RivinApr 25 '12 at 20:50

Presumably the generic Galois group is the group of order 1344, so these are the exceptional values where the Galois group becomes smaller.
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Noam D. ElkiesApr 25 '12 at 21:06

1

These are the cases when the polynomial is reducible.
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Mark SapirApr 26 '12 at 1:37

@Igor: Apologies, I should have qualified this with 'to the initial question'. @Mark: Of course, I can't believe I forgot to check something so obvious in my calculations!
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Alastair LitterickApr 26 '12 at 7:43

In arxiv.org/abs/1209.5300 I give the following polynomial with the same Galois group over Q(t): $(y+1)(y^7-y^6-11y^5+y^4+41y^3+25y^2-34y-29)-t(2y+3)^2$. This has $(6912t^4-3456t^3-95472t^2+23976t-1417)^2$ for a discriminant. This doesn't quite answer the second question, but substituting $y \mapsto (x-3)/2$, $t \mapsto s/256$ gives $(x-1) (x^7-23 x^6+181 x^5-547 x^4+515 x^3-325 x^2+71 x-1)-s x^2$, with discriminant $256 (27 s^4-3456 s^3-24440832 s^2+1571291136 s-23773315072)^2$, which does.